Answer :
Sure! Let's break down the problem step by step to evaluate [tex]\( 39.1 - 0.794 \)[/tex]:
1. Identify the numbers involved: We have the number 39.1, and we want to subtract 0.794 from it.
2. Setup the subtraction:
[tex]\[
39.1 - 0.794
\][/tex]
3. Perform the subtraction:
First, align the decimal points:
[tex]\[
\begin{array}{r}
39.100 \\
- 0.794 \\
\hline
\end{array}
\][/tex]
Next, subtract each digit starting from the rightmost digit (in the thousandths place):
- The thousandths place: [tex]\(0 - 4\)[/tex] (Since we can't subtract 4 from 0, we need to borrow from the next higher place.)
- Borrow 1 from the hundredths place: 10 - 4 = 6.
- The hundredths place: 9 - 9 = 0.
- The tenths place: 0 (borrowed from the whole number part) - 7 (since we are borrowing, we need to reduce the next significant digit)
Re-borrow from the units' place:
- Borrow 1 from 9 (part of 9 in the tens place):
[tex]\[ 10 - 7 = 3 \][/tex]
- The units place: 8 (borrowed from 3 in 39.1) - 0 = 8
Finally, the tens place: 38 remains unchanged since the original was 39, but we accounted 1 in 39 that was used for compensation.
4. Combine the results:
[tex]\[
\begin{array}{r}
39.100 \\
- 0.794 \\
\hline
38.306 \\
\end{array}
\][/tex]
The result of [tex]\( 39.1 - 0.794 \)[/tex] is [tex]\( 38.306 \)[/tex].
1. Identify the numbers involved: We have the number 39.1, and we want to subtract 0.794 from it.
2. Setup the subtraction:
[tex]\[
39.1 - 0.794
\][/tex]
3. Perform the subtraction:
First, align the decimal points:
[tex]\[
\begin{array}{r}
39.100 \\
- 0.794 \\
\hline
\end{array}
\][/tex]
Next, subtract each digit starting from the rightmost digit (in the thousandths place):
- The thousandths place: [tex]\(0 - 4\)[/tex] (Since we can't subtract 4 from 0, we need to borrow from the next higher place.)
- Borrow 1 from the hundredths place: 10 - 4 = 6.
- The hundredths place: 9 - 9 = 0.
- The tenths place: 0 (borrowed from the whole number part) - 7 (since we are borrowing, we need to reduce the next significant digit)
Re-borrow from the units' place:
- Borrow 1 from 9 (part of 9 in the tens place):
[tex]\[ 10 - 7 = 3 \][/tex]
- The units place: 8 (borrowed from 3 in 39.1) - 0 = 8
Finally, the tens place: 38 remains unchanged since the original was 39, but we accounted 1 in 39 that was used for compensation.
4. Combine the results:
[tex]\[
\begin{array}{r}
39.100 \\
- 0.794 \\
\hline
38.306 \\
\end{array}
\][/tex]
The result of [tex]\( 39.1 - 0.794 \)[/tex] is [tex]\( 38.306 \)[/tex].
